Posted by Christina on .
The mean number of bumped airline passengers per day is 1.35 and the standard deviation is 0.25. For a random selection of a group of 40 days, what is the probability that the mean of bumped passengers for the 40 days will be between 1.25 and 1.50?

stats. 
MathGuru,
Use this zscore formula for this problem:
z = (x  mean)/(sd/√n)
x = 1.25, 1.50
mean = 1.35
sd = 0.25
n = 40
Calculate two zscores, then use a ztable to determine probability between the two scores.
I hope this will help get you started. 
stats. 
Christina,
z=(1.251.35)/(0.25sqrt40)= 0.0791
z=(1.501.35)/(0325sqrt40)= 0.0791
but not sure how to read the z scores I don't know if it would be.5000.5000 or what can someone please help me on the z scores for 0.0791 and 0.0791 
stats. 
PsyDAG,
SD/√n = .25/6.3246 = .039528192 = .04
Z = (1.251.35)/.04 = .10/.04 = 2.5
Z = (1.501.35)/.04 = .15/.04 = ?
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of both Z scores between those Z scores and mean. Add the two together.